Chapter 2 Electrostatics: Problem 2.14

Problem 2.14 Find the electric field inside a sphere that carries a charge density proportional to the distance from the origin, $ρ = kr$, for some constant $k$. [Hint: This charge density is not uniform, and you must integrate to get the enclosed charge.]

Solution:

Given, $\rho = kr$ 

Consider a spherical Gaussian surface of radius $r_0$ such that $r_0\le R$ where $R$ is the radius of the sphere. 

 

 

 

 

 

 

 

 

 

 

 

By Gauss Law, $$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_0}$$ Since the charge distribution is radially symmetrical, the direction of the electric field will be radially outwards. $$\therefore E\times 4\pi {r_0}^2 = \frac{Q_{enc}}{\epsilon_0}$$

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$$Q_{enc} = \int_{vol.}\rho \times d\tau$$ $$Q_{enc} = \int_{vol.}kr \times r^2 sin\theta dr d\theta d\phi$$ $$Q_{enc} = k\int_0^{r_0} r^3 dr \int_0^{\pi} sin\theta d\theta \int_0^{2\pi}d\phi$$ $$Q_{enc} = k\times \frac{r^4}{4} \times 2 \times 2\pi$$ 

$$Q_{enc} = k\pi {r_0}^4$$

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$$\therefore E \times 4\pi r_0^2 = \frac{k \pi r_0^4}{\epsilon_0}$$

$$\therefore E = \frac{1}{4\epsilon_0}k{r_0}^2$$ 

$$\therefore E = \frac{1}{4\pi\epsilon_0}\pi k{r_0}^2$$

 $$\therefore \vec{E} = \frac{1}{4\pi\epsilon_0}\pi k{r_0}^2\hat{r}$$

 

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